leetcode 242. Effective ectopic letter word valid-anagram (Python3 achieve)

Title Description

Given two strings s and t, t write a function to determine whether the ectopic letters of the word s.

Example 1:

Input: s = "anagram", t = "nagaram"
Output: true
Example 2:

Input: s = "rat", t = "car"
Output: false
Description:
You can assume string contains only lowercase letters.

Topic solution

Method one: word frequency statistics to compare

class Solution:
    def isAnagram(self, s: str, t: str) -> bool:
        from collections import Counter
        maps = Counter(s)
        mapt = Counter(t)
        return maps == mapt

Method Two: Statistics letter No.

class Solution:
    def isAnagram(self, s: str, t: str) -> bool:
        c1 = [0]*26
        c2 = [0]*26
        
        for c in s:
            pos = ord(c) - ord('a')
            c1[pos] = c1[pos] + 1
            
        for c in t:
            pos = ord(c) - ord('a')
            c2[pos] = c2[pos] + 1
            
        return c1 == c2

Expand topics:

Source: https://www.v2ex.com/t/624125
questions asked: comparison of two words, the first look at whether a word after a word with the letters spliced out.

def can_be_composed(a, b):
    a_count = collections.Counter(a)
    b_count = collections.Counter(b)
    return all(a_count[k] <= b_count.get(k, 0) for k in a_count)

The same train of thought, word frequency is not just an equal sign, with <=, using all ensure that all conditions are met.